Question
If
\[ \cos^{-1}x > \sin^{-1}x \]
Find the range of \( x \).
Solution
Use identity:
\[ \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} \]
So inequality becomes:
\[ \cos^{-1}x > \frac{\pi}{2} – \cos^{-1}x \]
\[ 2\cos^{-1}x > \frac{\pi}{2} \]
\[ \cos^{-1}x > \frac{\pi}{4} \]
Now apply cosine:
\[ x < \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]
Also domain of x is:
\[ -1 \le x \le 1 \]
Hence,
\[ -1 \le x < \frac{1}{\sqrt{2}} \]
Final Answer:
\[ \boxed{-1 \le x < \frac{1}{\sqrt{2}}} \]
Key Concept
Convert inequality using identity and then apply monotonic property of cos⁻¹x.